Mathematics Solutions Solutions for Class 8 Maths Chapter 11 Division Of Polynomials are provided here with simple step-by-step explanations. These solutions for Division Of Polynomials are extremely popular among Class 8 students for Maths Division Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Divide. Write the quotient and the remainder.
(1) 21m2 ÷ 7m
(2) 40a3 ÷ (−10a)
(3) (−48p4) ÷ (−9p2)
(4) 40m5 ÷ 30m3
(5) (5x3 − 3x2) ÷ x2
(6) (8p3 − 4p2) ÷ 2p2
(7) (2y3 + 4y2 + 3) ÷ 2y2
(8) (21x4 − 14x2 + 7x) ÷ 7x3
(9) (6x5 − 4x4 + 8x3 + 2x2) ÷ 2x2
(10) (25m4 − 15m3 + 10m + 8) ÷ 5m3

(1) 21m2 ÷ 7m

$=\frac{21{m}^{2}}{7m}\phantom{\rule{0ex}{0ex}}=\frac{3×7×m×m}{7×m}\phantom{\rule{0ex}{0ex}}=3m$
So, quotient = 3m and remainder = 0

(2) 40a3 ÷ (−10a)
$=\frac{40{a}^{3}}{\left(-10a\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(-4\right)×\left(-10a\right)×{a}^{2}}{\left(-10a\right)}\phantom{\rule{0ex}{0ex}}=-4{a}^{2}$
So, quotient = $-4{a}^{2}$ and remainder = 0

(3) (−48p4) ÷ (−9p2)
$=\frac{-48{p}^{4}}{-9{p}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-48{p}^{2}×{p}^{2}}{-9{p}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{16{p}^{2}}{3}$
So, quotient = $\frac{16{p}^{2}}{3}$ and remainder = 0.

(4) 40m5 ÷ 30m3
$=\frac{40{m}^{5}}{30{m}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{4×10{m}^{3}×{m}^{2}}{3×10{m}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{4{m}^{2}}{3}$
So, quotient = $\frac{4{m}^{2}}{3}$ and remainder = 0.

(5) (5x3 − 3x2) ÷ x2
$=\frac{5{x}^{3}-3{x}^{2}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}\left(5x-3\right)}{{x}^{2}}\phantom{\rule{0ex}{0ex}}=5x-3$
So, quotient = $5x-3$ and remainder = 0

(6) (8p3 − 4p2) ÷ 2p2
$=\frac{8{p}^{3}-4{p}^{2}}{2{p}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2{p}^{2}\left(4p-2\right)}{2{p}^{2}}\phantom{\rule{0ex}{0ex}}=4p-2$
So, quotient = $4p-2$ and remainder = 0

(7) (2y3 + 4y2 + 3) ÷ 2y2
$=\frac{2{y}^{3}+4{y}^{2}+3}{2{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2{y}^{2}\left(y+2\right)+3}{2{y}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2{y}^{2}\left(y+2\right)}{2{y}^{2}}+\frac{3}{2{y}^{2}}\phantom{\rule{0ex}{0ex}}=\left(y+2\right)+\frac{3}{2{y}^{2}}$
So, quotient = $\left(y+2\right)$ and remainder = 3

(8) (21x4 − 14x2 + 7x) ÷ 7x3
$=\frac{21{x}^{4}-14{x}^{2}+7x}{7{x}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{7{x}^{3}×3x-14{x}^{2}+7x}{7{x}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{7{x}^{3}×3x}{7{x}^{3}}+\frac{-14{x}^{2}+7x}{7{x}^{3}}\phantom{\rule{0ex}{0ex}}=3x+\frac{-14{x}^{2}+7x}{7{x}^{3}}$
So, quotient = 3x and remainder = $-14{x}^{2}+7x$

(9) (6x5 − 4x4 + 8x3 + 2x2) ÷ 2x2
$=\frac{6{x}^{5}-4{x}^{4}+8{x}^{3}+2{x}^{2}}{2{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2{x}^{2}\left(3{x}^{3}-2{x}^{2}+4x+1\right)}{2{x}^{2}}\phantom{\rule{0ex}{0ex}}=3{x}^{3}-2{x}^{2}+4x+1$
So, quotient = $3{x}^{3}-2{x}^{2}+4x+1$ and remainder = 0

(10) (25m4 − 15m3 + 10m + 8) ÷ 5m3
$=\frac{25{m}^{4}-15{m}^{3}+10m+8}{5{m}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{5{m}^{3}\left(5m-3\right)+\left(10m+8\right)}{5{m}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{5{m}^{3}\left(5m-3\right)}{5{m}^{3}}+\frac{\left(10m+8\right)}{5{m}^{3}}\phantom{\rule{0ex}{0ex}}=\left(5m-3\right)+\frac{\left(10m+8\right)}{5{m}^{3}}$
So, quotient = $5m-3$ and remainder = $10m+8$

#### Question 1:

Divide and write the quotient and the remainder.
(1) (y2 + 10y + 24) ÷ (y + 4)
(2) (p2 + 7p − 5) ÷ (p + 3)
(3) (3x + 2x2 + 4x3) ÷ (x − 4)
(4) (2m3 + m2 + m + 9) ÷ (2m − 1)
(5) (3x − 3x2 − 12 + x4 + x3) ÷ (2 + x2)
(6) (a4a3 + a2a + 1) ÷ (a3 − 2)
(7) (4x4 − 5x3 − 7x + 1) ÷ (4x − 1)

(1) (y2 + 10y + 24) ÷ (y + 4) So, quotient = y + 6 and remainder = 0

(2) (p2 + 7p − 5) ÷ (p + 3) So, quotient = p + 6 and remainder = $-$17.

(3) (3x + 2x2 + 4x3) ÷ (x − 4) So, quotient = $4{x}^{2}+18x+75$ and remainder = 300.

(4) (2m3 + m2 + m + 9) ÷ (2m − 1) So, quotient = ${m}^{2}+m+1$ and remainder = 10.

(5) (3x − 3x2 − 12 + x4 + x3) ÷ (2 + x2) So, quotient = ${x}^{2}+x-5$ and remainder = x $-$ 2.

(6) (a4 − a3 + a2 − a + 1) ÷ (a3 − 2) So, quotient = $a-1$ and remainder = ${a}^{2}+a-1$.

(7) (4x4 − 5x3 − 7x + 1) ÷ (4x − 1) So, quotient = ${x}^{3}-{x}^{2}-\frac{x}{4}-\frac{29}{16}$ and remainder = $-\frac{13}{16}$.

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